Lorentz scalar

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In physics a Lorentz scalar is a scalar which is invariant under a Lorentz transformation. A Lorentz scalar is generated from vectors and tensors. While the vectors and tensors are altered by Lorentz transformations, scalars are unchanged.

1 Simple scalars in special relativity
2 Energy, rest mass, 3-momentum, and 3-speed from 4-momentum
3 More complicated scalars
4 See also
5 References

[edit] Simple scalars in special relativity

[edit] The length of a position vector

World lines for two particles at different speeds.

In Special relativity the location of a particle in 4-dimensional spacetime is given by its world line

$x^{\mu} = (ct, \mathbf{x} )$

where $\mathbf{x} = \mathbf{v} t$ is the position in 3-dimensional space of the particle, $\mathbf{v}$ is the velocity in 3-dimensional space and $c$ is the speed of light.

The "length" of the vector is a Lorentz scalar and is given by

$x_{\mu} x^{\mu} = \eta_{\mu \nu} x^{\mu} x^{\nu} = (ct)^2 - \mathbf{x} \cdot \mathbf{x} \ \stackrel{\mathrm{def}}{=}\ \tau^2$

where $τ$ is c times the proper time as measured by a clock in the rest frame of the particle and the metric is given by

$\eta^{\mu\nu} =\eta_{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$ .

This is a time-like metric. Often the Minkowski metric is used in which the signs of the ones are reversed.

$\eta^{\mu\nu} =\eta_{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$ .

This is a space-like metric. In the Minkowski metric the space-like interval $s$ is defined as

$x_{\mu} x^{\mu} = \eta_{\mu \nu} x^{\mu} x^{\nu} = \mathbf{x} \cdot \mathbf{x} - (ct)^2 \ \stackrel{\mathrm{def}}{=}\ s^2$ .

We use the Minkowski metric in the rest of this article.

[edit] The length of a velocity vector

The velocity vectors in spacetime for a particle at two different speeds. In relativity an acceleration is equivalent to a rotation in spacetime

The velocity in spacetime is defined as

$v^{\mu} \ \stackrel{\mathrm{def}}{=}\ {dx^{\mu} \over d\tau} = \left (c {dt \over d\tau}, { dt \over d\tau}{d\mathbf{x} \over dt} \right ) = \left ( \gamma , \gamma { \mathbf{v} \over c } \right )$

where

$\gamma \ \stackrel{\mathrm{def}}{=}\ { 1 \over {\sqrt {1 - {{\mathbf{v} \cdot \mathbf{v} } \over c^2} } } }$ .

The magnitude of the 4-velocity is a Lorentz scalar and is minus one,

v μ v μ = - 1

The 4-velocity is therefore, not only a representation of the velocity in spacetime, is also a unit vector in the direction of the position of the particle in spacetime.

[edit] The inner product of acceleration and velocity

Diagram 1. Changing views of spacetime along the world line of a rapidly accelerating observer.

In this animation, the dashed line is the spacetime trajectory ("world line") of a particle. The balls are placed at regular intervals of proper time along the world line. The solid diagonal lines are the light cones for the observer's current event, and intersect at that event. The small dots are other arbitrary events in the spacetime. For the observer's current instantaneous inertial frame of reference, the vertical direction indicates the time and the horizontal direction indicates distance.

The slope of the world line (deviation from being vertical) is the velocity of the particle on that section of the world line. So at a bend in the world line the particle is being accelerated. Note how the view of spacetime changes when the observer accelerates, changing the instantaneous inertial frame of reference. These changes are governed by the Lorentz transformations. Also note that:
• the balls on the world line before/after future/past accelerations are more spaced out due to time dilation.
• events which were simultaneous before an acceleration are at different times afterwards (due to the relativity of simultaneity),
• events pass through the light cone lines due to the progression of proper time, but not due to the change of views caused by the accelerations,and
• the world line always remains within the future and past light cones of the current event.

The 4-acceleration is given by

$a^{\mu} \ \stackrel{\mathrm{def}}{=}\ {dv^{\mu} \over d\tau}$ .

The 4-acceleration is always perpendicular to the 4-velocity

$0 = {1 \over 2} {d \over d\tau} \left ( v_{\mu}v^{\mu} \right ) = {d v_{\mu} \over d\tau} v^{\mu} = a_{\mu} v^{\mu}$ .

Therefore, we can regard acceleration in spacetime as simply a rotation of the 4-velocity. The inner product of the acceleration and the velocity is a Lorentz scalar and is zero. This rotation is simply an expression of energy conservation:

${d E \over d\tau} = \mathbf{F} \cdot { \mathbf{v} \over c}$

where $E$ is the energy of a particle and $\mathbf{F}$ is the 3-force on the particle.

[edit] Energy, rest mass, 3-momentum, and 3-speed from 4-momentum

See [Ref. 2, P. 65]. A space-like metric is used.

The 4-momentum of a particle is

$p^{\mu} = m v^{\mu} = \left ( \gamma m , \gamma { m \mathbf{v} \over c } \right ) = \left ( \gamma m , { \mathbf{p} \over c } \right ) = \left ( {E \over c^2 } , { \mathbf{p} \over c } \right )$

where $m$ is the particle rest mass, $\mathbf{p}$ is the momentum in 3-space, and

E = γ m c 2

is the energy of the particle.

[edit] Measurement of the energy of a particle

Consider a second particle with 4-velocity $u$ and a 3-velocity $\mathbf{u}_2$ . In the rest frame of the second particle the inner product of $u$ with $p$ is proportional to the energy of the first particle

$p_{\mu} u^{\mu} = - { E_1 \over c^2}$

where the subscript 1 indicates the first particle.

Since the relationship is true in the rest frame of the second particle, it is true in any reference frame. $E 1$ , the energy of the first particle in the frame of the second particle, is a Lorentz scalar. Therefore

${ E_1 \over c^2} = \gamma_1 \gamma_2 m_1 - \gamma_2 \mathbf{p}_1 \cdot \mathbf{u}_2$

in any intertial reference frame, where $E 1$ is still the energy of the first particle in the frame of the second particle .

[edit] Measurement of the rest mass of the particle

In the rest frame of the particle the inner product of the momentum is

$p μ p μ = - m 2$ .

Therefore $m 2$ is a Lorentz scalar. The relationship remains true independent of the frame in which the inner product is calculated.

[edit] Measurement of the 3-momentum of the particle

Note that

$\left ( p_{\mu} u^{\mu} \right )^2 + p_{\mu} p^{\mu} = { E_1^2 \over c^4} -m^2 = \left ( \gamma_1^2 -1 \right ) m^2 = \gamma_1^2 { {\mathbf{v}_1 \cdot \mathbf{v}_1 } \over c^2 }m^2 = \mathbf{p}_1 \cdot \mathbf{p}_1$ .

The square of the magnitude of the 3-momentum of the particle as measured in the frame of the second particle is a Lorentz scalar.

[edit] Measurement of the 3-speed of the particle

The 3-speed, in the frame of the second particle, can be constructed from two Lorentz scalars

$v_1^2 = \mathbf{v}_1 \cdot \mathbf{v}_1 = { { \mathbf{p}_1 \cdot \mathbf{p}_1 c^6 } \over { E_1^2 } }$ .

[edit] More complicated scalars

Scalars may also be constructed from the tensors and vectors, from the contraction of tensors, or combinations of contractions of tensors and vectors.